 Solution

A deformation maps a material point in the reference body to its current position, $y:\Omega \to {\mathbb{R}}^3$, with deformation gradient $F=Dy$. A homogeneous deformation has constant gradient and is affine, $y(X)=FX+b$.

If $\det F>0$, the matrix $F^{T}F$ is symmetric positive definite. The matrix-square-root theorem says that any such matrix has a unique symmetric positive definite square root, obtained by taking positive square roots in its orthogonal spectral decomposition. SetThen $R^{T}R=U^{-1}{F}^{T}F{U}^{-1}=I$ and $\det R=\det F/\det U>0$, so $R\in \mathrm{SO}(3)$ and $F=RU$. If also $F=\tilde{R}\tilde{U}$, then $F^{T}F={\tilde{U}}^2$; uniqueness of the positive square root gives $\tilde{U}=U$ and then $\tilde{R}=R$. This proves existence and uniqueness of polar decomposition.

The principal stretches are the positive eigenvalues of $U$, equivalently the singular values of $F$ or square roots of the eigenvalues of $F^{T}F$.